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Thread: going in circles

  1. #1
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    Question going in circles

    Hi folks,

    I am trying to solve the following question:

    Find the equations of the two circles which pass through the point (2,0) and have both the y-axis and the line y - 1 = 0 as tangents.

    This seems like an easy question because you can do it by scale drawing. The first (small) circle can't really go anywhere else - see attachment! But assuming you had to do it by calculation.

    To find the equation of a circle we need 3 points on the circumference. We have one supplied directly (2,0) then we have the general equation:

    $\displaystyle
    x^2 + y^2 + 2fx + 2gy + c = 0 (1)
    $

    plus the two tangents:

    x = 0

    y = 1

    substituting (2,0) gives:

    4 + 4f + c = 0

    and substituting x = 0 into the general equation gives:

    $\displaystyle y^2 + 2gy -4(f + 1) = 0$

    a quadtratic in y. Since the tangent touches the circle we expect a repeated root i.e. $\displaystyle b^2 - 4ac = 0$

    $\displaystyle 4g^2 + 4.1.4(f + 1) = 0$

    $\displaystyle g^2 + 4f + 4 = 0 (4)$


    then substitute y = 1 into the general equation:

    $\displaystyle x^2 + 1 + 2fx + 2g -4(f + 1) = 0$

    $\displaystyle x^2 + 2fx + 2g + 1 - 4(f + 1) = 0$

    again setting $\displaystyle b^2 - 4ac = 0 $ gives:

    $\displaystyle
    4f^2 + 16f - 8g + 12 = 0 (5)
    $


    Equations (4) and (5) are correct. What I mean is that when you substitue the correct answers in, they give consistent results but how do you solve them? It seems clear that I am on the wrong track. I then realised that the equation y = -x + 1 must be a diameter to both circles so I tried substituting this into the general equation (1) to solve for the intersection points. It yielded a result for x with many terms in f and g that I couldn't reduce to anything like the correct answer.

    Any help here would be appreciated!
    best regards
    Attached Thumbnails Attached Thumbnails going in circles-m16f7.bmp  
    Last edited by s_ingram; Jul 23rd 2009 at 01:50 AM.
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  2. #2
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    Hi

    AS you said the line y=-x+1 is a diameter of the circles
    The centers are on this line
    Let C (xC,yC) one center of the possible circles
    yC = -xC + 1

    The distance between C and the point M(2,0) is such that CM² = (xC-2)²+yC² = 2xC² - 6 xC + 5
    This distance is the same as the one between C and the y-axis, which is |xC|
    Therefore : 2xC² - 6 xC + 5 = xC²
    xC² - 6 xC + 5 = 0
    (xC - 1)(xC - 5) = 0
    xC = 1 or 5
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  3. #3
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    Quote Originally Posted by s_ingram View Post
    Hi folks,

    I am trying to solve the following question:

    Find the equations of the two circles which pass through the point 2,0) and have both the y-axis and the line y-1=0 as tangents.

    ...
    1. I used this equation of a circle:

    $\displaystyle (x-a)^2+(y-b)^2=r^2$ with the center at C(a, b) and the radius r.

    2. You already know 3 points of the circle: (2, 0), (0, 1-r), (a, 1)

    3. Substitute the coordinates of these points into the equation of the circle. You'll get a system of equations. Solve this system for a, b and r.

    $\displaystyle \left|\begin{array}{rcl}(2-a)^2+(0-b)^2&=&r^2 \\ (0-a)^2+(a+1-b)^2&=&r^2 \\ (a-a)^2 + (1-b)^2&=&r^2 \end{array}\right.$

    4. After some transformations and calculations you'll get:

    Circle #1: $\displaystyle (1,0); r=1$
    Circle #2: $\displaystyle (5,-4); r=5$
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  4. #4
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    Hello, s_ingram!

    Find the equations of the two circles which pass through the point (2,0)
    and have both the $\displaystyle y$-axis and the line $\displaystyle y\,=\, 1$ as tangents.
    Code:
            |
            |         R
          - + - P - * o * - - - y=1
        ----+---o-----:-----*----------
            | * 2     :       *
            |*        :        *
            |         :
            *         :         *
          Q o - - - - o C       *
            *       (x,y)       *
            |
            |*                 *
            | *               *
            |   *           *
            |       * * *
            |

    The center of the circle is $\displaystyle C(x,y)$

    It passes through the point $\displaystyle P(2,0)$
    It is tangent to the $\displaystyle y$-axis at $\displaystyle Q(0,y)$
    It is tangent to $\displaystyle y = 1$ at $\displaystyle R(x,1)$

    The radius is: .$\displaystyle CP \:=\: CQ \:=\: CR \:=\: r$


    We have: .$\displaystyle x \:=\:r$

    Since $\displaystyle CR \:=\:1-y \:=\:r\:=\:x$, then: .$\displaystyle y \:=\:1-x$

    Since $\displaystyle CP \,=\,R$, we have: .$\displaystyle \sqrt{(x-2)^2 + (y-0)^2} \:=\:r$

    . . which becomes: .$\displaystyle \sqrt{(x-2)^2 + (1-x)^2} \:=\:x$

    Then: .$\displaystyle (x-2)^2 + (1-x)^2 \:=\:x^2$

    . . which simplifies to: .$\displaystyle x^2 - 6x + 5 \:=\:0$

    . . which factors: .$\displaystyle (x-1)(x-5) \:=\:0$

    . . and has roots: .$\displaystyle x \:=\:1,5$

    . . . . . . . . Then: .$\displaystyle y \:=\:0,\text{-}4$


    One circle has center $\displaystyle (1,0)$ and $\displaystyle r = 1\!:\quad \boxed{(x-1)^2 + y^2 \:=\:1}$

    The other has center $\displaystyle (5,\text{-}4)$ and $\displaystyle r = 5\!:\quad \boxed{(x-5)^2 + (y+4)^2 \:=\:25}$

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  5. #5
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    Excellent! Thanks a lot guys.
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